# modus ponens

• Jun 30th 2010, 08:01 PM
dunsta
modus ponens
premise is -p

-r -> p therefore -r, modus ponens

is this assumption correct?
• Jun 30th 2010, 08:06 PM
Jhevon
this is not using modus ponens, and it's not true anyway. Remember, \$\displaystyle \neg R \implies P\$ is the same as saying \$\displaystyle \neg P \implies R\$. Since \$\displaystyle \neg P\$ is true, \$\displaystyle R\$ follows by modus ponens, not \$\displaystyle \neg R\$
• Jun 30th 2010, 08:28 PM
dunsta
thanks
premise -q
-p or q, therefore p, disjunctive syllogism

if q is false then p must be true, otherwise the whole condition is false??

or is it,
if -q(i.e. q is false)
then -p must be true for -p or q.

??
so
premise -q
-p or q, therefore -p, disjunctive syllogism

??
• Jun 30th 2010, 10:26 PM
Jhevon
Quote:

Originally Posted by dunsta
so
premise -q
-p or q, therefore -p, disjunctive syllogism

??

correct
• Jun 30th 2010, 11:32 PM
dunsta
thanks for the help

the question I am working on is:

-r --> p, (-p) or q, -s -->(-p)and (-r), (-p) and r --> (-s) or t, -q, therefore t.

So far I have
1: -p, -p or q therefore -p conjunctive syllogism
2: -p, -r --> p therefore r modus ponens
3: -p, r, therefore -p and r conjunctive addition
(the next 2 lines I am unsure of)
4: -p and r, -s --> (-p) and(-r) /*-p is true from 1. but r is true from 2, so -r is not true. therefore (-p) and(-r) is not true so - (-p) and(-r) = -(p or r) demorgans.??? I don't know where to go with this? */
therefore s, because -s --> (-p) and(-r) is false.
5: (-p) and r (from 3), -s is false, (-s) or t therefore t disjunctive syllogism

am I close to showing the argument is valid??
thanks for any input and help
so -s --> (-p) and(-r)